Question

An ideal gaseous mixture of ethane (C2H) and ethene (CH2) occupies 28 litre at latm, 0°C. The mixturereacts completely with 128 gm O, to produce CO, and H,O. Mole fraction at C,H, in the mixture is-(A) 0.6(B) 0.4(C) 0.5(D) 0.8​

Answer 1

Answer : The mole fraction of ethane and ethene in gaseous mixture are, 0.5 and 0.5

Explanation :

First we have to calculate the moles of gaseous mixture of ethane and ethene.

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 28 L

T = temperature of gas = $0^oC=273+0=273K$

n = number of moles of gaseous mixture = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (28L)=n\times (0.0821L.atm/mol.K)\times (273K)$

$n=1.249mole$

Thant means,

$n_{C_2H_6}+n_{C_2H_4}=1.249$

Let the moles of $C_2H_6$ and $C_2H_4$ be, 'a' and 'b' respectively.

$a+b=1.249$        ............(1)

The balanced chemical reactions are:

(a) $C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

(b) $C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

From both the reaction, we conclude that

$\frac{7a}{2}+3b=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}$

$\frac{7a}{2}+3b=\frac{130}{32}$       ...........(2)

Now by solving the equation 1 and 2, we get:

a = 0.631

b = 0.618

Now we have to calculate the moles fraction of ethane and ethene in gaseous mixture.

$\text{Mole fraction of }C_2H_6=\frac{\text{Moles of }C_2H_6}{\text{Moles of }C_2H_6+\text{Moles of }C_2H_4}=\frac{0.631}{0.631+0.618}=0.505\approx 0.5$

$\text{Mole fraction of }C_2H_4=\frac{\text{Moles of }C_2H_4}{\text{Moles of }C_2H_6+\text{Moles of }C_2H_4}=\frac{0.618}{0.631+0.618}=0.495\approx 0.5$

Therefore, the mole fraction of ethane and ethene in gaseous mixture are, 0.5 and 0.5

Answer 2

Answer:

0.5 is the answer

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